1.7 Knowing that the central portion of the link BD has a unifonn cross-sectional area of 800 mm2, detennine the magnitude of the load P for which the nonnal stress in that portion of BD is 50 MPa. Problem 1.7 OBI> -= 50 M Po.. ::- So ~ 10' <7 2. 0 eD ':" 000 W\W\ =- oDO)( A ~Ct 10 -,:l- WI Fao = oSDABD= (.sO></O")(/?oox/6() D",o.,.;+iI'H- bootl ,L.~V'4.~ Cj t c,. Clfbod, +, Me.= FQD p=- (S'JO =- ~D ) - 0.135 P ': 0 I.SC.'6b FaD 'P ':"(1.5b8b p A'B D. 0 240 (O.lfSo) = qox/O" tv )(4-U)( IO!) c.;2.7)( 103. N r;;'2.7 ldJ -lilt Problem 1.10 1.10 Two horizontal 5-kip forces are applied to pin B of the assemblyshown. Knowing that a pin of D.8-in.diameter is used at each connection, determinethe maximumvalue of the averagenormalstress (a) in linkAB, (b) in linkBC. Use jDi\l\t B C<..$-tV'ee boJy. 10 K",s F~e~< fO I<.'ps L<.1'V of Foil'/: t +r-I'GlII\~Ie. S iVles FAa &. ,'", 600 - s,'." 'ISO FA-G ~ AS LI'~ k S -- - ,.'Iv orb 051.., ~. F~c:. -= 7. '32 DS k"f'5 a'S Q.. +elf\~jo ~ e~ (el) stress .;t1A8 Lank BC ,'s Q. ~()N'\pr--e-s.siC»"l c.~"ss (6) s e.c+i()V\~ S1¥es.s i~ 8(' q,v-eet p,"" ,'s Gee. .:: ' a. O.S'II\ lA\rla = (I. g - 0.8 )(O.£") s eCJh' o 6:Aa-- kIps. be-" M'V'lIIJ"'" t:tf 8.QbS8 P"'B .. Aliff' .- - 7;.5:(0'5"" :: 0.5 Iif . f;;lf '<5" ... ~e~beV' A: (f. S ) ( O. 5) = - Fee. - A - o. q "" '2. - 3. Cf~S'8 =- - cr.q6 k.st o. <:t ~ Problem 1.17 0.4 in. 1.17 A load P is applied to a steel rod supported as shown by an aluminum plate into which a O.6-in.-diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, detennine the largest load P that may be applied to the rod. --1 0.25 in. t Fo~ Fot" , ~..Pv ,;nv",,- rt2. = ~2. LiMI'+"v\~ P 1:' ':. A. p :. v~jve A2. = iT" d AI = steel ... "lI t =- -IT ( = ?: P IS tJ,e .IIII(0. ~)(0. '.n L D. ;SlfD A, 1:'. ~ (0. 7.No¥~) == 13.57 k"ps ,.<;) ( C>.~ 5') 1> = A2.1::2.= (I..'-Sbb )('0-) (.)f 1T r.lt = 11 =- I. :zSbb i",1. -;: \ <. S"7 k;f-S 5t'11e~jJe.l"'VCA-JlJe'" P '::" ':<.SI k'I':).5 .... Problem 1.31 1.31 The 1.4 kip load P is supported by two wooden members of uniform cross sectionthat arejoined by the simplegluedscarfspliceshown. Determinethenormal and shearingstressesin the glued splice. p::- At:.:- {s.O 6= e 14-00 110 P )(3.0) GOS ~ ~ e A, ::- 9()t::> - bO. -= 3Do IS 1i'\1. (/'-/DO )((..0$ 30")2. - Is D -= 70.0 p.,.t' ..... T-= P:s;",':?e 2Ao - (I '-tOO '\ s;V\ 60" (2J( IS> L =- 4o_4 psi ..... Problem 1.34 1.34 A steel pipe of300-mm outer diameter is fabricated nom 6-mm-thick plate by welding along a helix that forms an angle of 25° with a plane perpendicular to the axis of the pipe. Knowing that the maximum allowable normal and shearing stresses in the directions respectively normal and tangential to the weld are u= 50 MPa and .= 30 MPa, determine the magnitude P of the largest axial force that can be applied to the pipe. p 6mm -Ir0.300 dlJ: weld~ I'll r..:= to - t:: fo:: 1'010:: O.ISo - 0.00.6 = O.It.ft1 Ao ':. 1f ('f'01..- r~~):: Ba.seJ \G" \ 0'" p::- :: 50 Aors Ct>& 1 e Bc...seJ o~ l't \ = 30 p = ~ Ao 1:: ~h SY'f'IalJet/' vAlve MPo.: = ( S".~ ~e IS. +~~ ~/O-g Vt'i2.. ;l.So :: (5:: ro CDSze ><I Q. ~ ) (Sc X' (96 ') e.os'Z. 25'0 M~ ::: ~ rr ( (). ISO7..- 0.. I"+t.i 2.) ::::- S_StJ e 0. \5"0 M ?:. :: ;5;0 - 337 , ></03 :: 4-34 X 10 '51'", 28 (Z)(s..SL/~o-~)(30xlO6) :3 si... 500 JjJo",,~4e vc...Jv~ of "P I" P = 331 kN -' ..~ 1.35 A 240-kip load P is applied to the granite block shown. Determine the resulting maximum value of (a) the normal stress, (b) the shearing stress. Specify the orientation of the plane on which each of these maximum values occurs. Problem 1.35 p 36 AD : (,)( Eo) G:::: (0...) p - AQ CO$ t'I'1c:;..,6te", 5"; M~.. ( b) 2.8 . l. rh -2 6'to s-h..-e..s. ~o~t-e-~si".e. ~t Cu ==. - '.67 Co):''1.8 o.-t e = 90° :;: 0 ~+V"'e5.$ = <0.67 ks,. e= e ~ 0'" 'T..."1'::' ~~o ::- (~~ ~~6) ~+ 'a.S := ~ LfSo 3. ~3 K'=>" ~ 1.46 Two wooden members of 3.5 x 5.5-in. unifonn rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 75 psi, detennine the largest axial load p that can be safely applied. Problem 1.46 5.5 in. = (3.S 1\0 e r -::::.f 'P = 'P :51'..ec.o.sG 2A.~ S \-'1~e '=: '10° -= :<1\0 S . )(5.5) - 20° - =- I Cf.?5 j",~ 70" 'J I Y'I .-{ B (:{)( Icr. 2 ~ )L7§) sin 14Do -= 44'11- jib 4. 41 /<;r.s PROPRIETARY MATERIAL. @ 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. -4 Problem 1.51 1.51 Link A C is made of a steel with a 65-ksi ultimate normal stress and has a t -in. uniform rectangular cross section. t x It is connected to a support at A and to f memberBCD at C by -in.-diameterpins, while member BCD is connectedto its support at B by a I~-in.-diameter pin; all of the pins are made of a steel with a 25-ksi ultimate shearing stress and are in single shear. Knowing that a factor of safety of 3.25 is desired, determine the largest load P that can be applied at D. Note that link AC is not reinforced around the pin holes. Bin. D Use +v-ee bo,ty +J p MB p ~< B,r-4f = 0: (G)Ci~ FAc.) - -=- o. Lf~ +) Mc ::, 0: 2 B.j:- is: + ~ 'B.J"). -: -/1. ';.5,1. -+ (~)~ She"-'" j", p;"s L"" -= r,~c. oVlnet F~~ =- S~.» eo/' vJve Fo"o.>~ (,) S~eo."" in B= ::= -y", F:~ .sed-I'C)"", 6' A-t = :II oJ7. :: tt ev+ Ff'u"- (:t) AP)(.)..,j,JJ~ . P"" P . P I.e. 1.&.11'-(,.1 - 0 = 0 ~ - G B:f - 4 P A C~s ':ill.V'3 )' ':3.~~) 4 k-s ~ '" 2. I .3 'P iI p = 0.7058'-6 P '8 (2) =- D.1?<t~ 5' k;rs <:Uo\",,1 C. ~ o. 6~.S k:rs O.l:2.S k:ps . p -= (O.I..f~)(0.'2.'5') 'L A ~ ~F: 10 Ae. '- Zs P = ~~, Ah.t ~ (3~~-rtit Xi.- ;) of' F,.<.. is p;'" J B- A c..""J Co.. . API'" 1:: Te.VI'5I'OV\ wt p -.3 .,.. 0 Ct~ . X B)( =- I~ ~c tIp 10 P I=AC. -+21=,,= r 0: By I B BCD. = o. '300 k;rs 13. ~ t 1=:s. .!LJL~ ~ ~S\l1[ -= (O.70S-8:8)(O.S8q~Cj) vJve of' 'P ;$ f~'l S' ( 3. :<5' 1\ &f)( 1C:.) Ly '=' 0.4/' OS-gqqq . k: P~ k:ps P =: 5Y>1e..U-et/' V".,YIJe,. 6~ 0.300 p -= 300 k:ps l.h .... Problem 1.55 1.55 In the steel structure shown, a 6-mm-diameter pin is used at C and IO-mmdiameter pins are used at B and D. The ultimate shearing stress is ISO MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes. 18mm + ee Use. D Front view +'LM(:,. 6 mm D.28b A 'I B ~w~'-~w bod, A Be.': 0: p - O.I~O Fe.o = C> (I) ~ Fa., p"" Side view +J~HI3 p =0: 'oomm['oomm1 o. I(.0 P A B Top view p C - O.':;( 0 C- ,3. ~ ,~C = 0 (~) JeV\'5io,,", 01'\ net- sed-;oV\ of J>;V\kB D. ~GOf Gt p1 ~C1O-= 6" A~t -:: ~. AMf =- (400)((O)(6)iIO-S)(I~-IO)(lo3) 3 l:.JtO -= SheCti" ;VI ry Fs.D $w-J pi == v\s d 8 . ::> Api'" 1:<.1 (1 ) SheaV' C ;V\ :: lJ. F.S. 't Pe". v,.}!)£? of' ~ f"O"'" CtVlJ ('1.') SMJJ e'" vJVf? N d 2. -- ( Jso')l./O" )(lL4 )(, 0 2, . l >l -3 0 )2 - 1>.9~~O: ,>,/03 N Feo IS 5.'91.,,70.></0'3N.. pi \11 c.J Wvw-. ,o? '[). I.G~.3 ~ /03 N 1> = (~)( Bll:f J..7ox[!cl) -: 2 rC Api... )l =- C ~ ~. ~c1?.'" (2)('5"0 ;(0'- 'P = (~Y1.g;('N'Ic:{d) :- )(~ )(b)C/O~)a.:" .Z.~~7'f .,./o'J. N 2. ,~)( lo~ tJ ~f -p i~ JJo..J ~Uf(' vc..L~" 'P =- 1.'~3 x 10' tV 'P"'" I.' 83 IetJ ~ 1.65 The 2000-lb load may be moved along the beam BD to any position between stops at E and F. Knowing that a.1I= 6 ksi for the steel used in rods AB and CD, determine where the stops should be placed if the permitted motion of the load is to be as large as possible. Problem 1.65 60in. P~t" "",dt~J AS ~e- be".. (SJ... +o.rc..e..s: = CD: '-J7lo Fc.,p t-- 10 ~~ 60;... p= ~~)h Use. Me- he'" ib +:> L HD - (60 =- )( )(i) I. 'tD7g k:f'.s 4. f H. 2.DOO k:rs .., + (60 Go - x:.. -= b0 P E? =- - 'X.,E)P FAB ~ '):SF=- = 0 (bo '>( I. 17g Joj 2. 0 0 0 35.3'f3 ~G Go ~~D- bo...l7" e. 0 ) FAB = )(;i) kps BE F D 4.S 'P = A.OOQ +)ZHg )( (F'"D ),.,"'" -= OJR Ac.p = (b ':" t='AIJ = (6 =- OJ.( A -= '2'1.7 iYl. .4 0 X-P bO ~D P = 0 "=' (bo)(J.g4o7i,) Z. 000 X-p;:" 55.:< in. ~ Problem 1.67 1.67 Each of the two vertical links CF connecting the two horizontal members AD and EG has a lOx 40-mm unifonn rectangular cross section and is made of a steel with an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20-mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa. Detennine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. 1"250 mm 1---400m~ B I 250 ml Use MeMbel" EiFG 0..$ ~ee bcJy- t Fee F ~ ~F O.Lfo ---+=-°.25"' 24kN 21l'kN 1) ~ ME = 0 O.LfC Fc'F"-(O.G5')(~'t""O!) Fe.F ':: 8a.sed in .Pin ks 0111 te.~5ili¥l Cb- d) t A::: on = (0.0£10 - 0.02.)(0.010) &o!} I-::;Ji? .sheC(.~;n A '= ~r;l F", ~ Ar':.+,}(,v .JJ.. t~ 2 YuA F,) 7T = ft(o.O2o} := tS {:"Q.(;,-L;)If' ot -se:o:fe.ty IO~ N ~ = '20D t./o-c \"1'1" (()~ej"nk) ':" IGO.O "1l103 N P,"YlS = 314.'bx/O -, ~ ~ (2){t5D><ID')(3ILf." )C/o-') = <7't.248 )(/O! SW1a.JJev'" \le.J~e.) F;S. ~ 0 CF Fu '=25'vA= (2(400'tIO')('200'JC/O-~) B~sed 3Cf)( c ~IJ ': '-CI=" i.e.. Ft) =- QI/.'248 'ftf.'248 :/0'3 3'" ~ 10 ~/O3 ': N N Z.1.f2 .. Problem 1 .69 1.69 The two portions of member AB are glued together along a plane forming an angle 0 with the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine (a) the value of 0 for which the factor of safety of the member is maximum, (b) the corresponding value of the factor of safety. (Hint: Equate the expressions obtained for the factors of safety with respect to normal stress and shear.) Ao :: (:?D)(J. 2..5) 1:: ';..~b ia-t1- A-t rl..e °p+,'V!f\v"",, o.YI~Je 0=:5')0- -=(FS.)t' ,,/ n 'r.::p-, 'l/,\ . "D f5'uA.:> IV,O~ ClA' S V'"€.$.S' '-.J = - C<:>.$ t::7 '. IV <S"= L + ( rs -: ) r.. . \) Sh e.Ct"':Vltl\ sfV' e<;$: ..J 1:'.;:- 'P.s Ao I", e (.OS ( r-t-. E1vJ;k: h ' S~J'II\ V\ Gb) ~ ~: A0 'P~ = GuAo 'P e -- ('" '\ ;:0. It: ,: 'P pI,) I 1:' = 'PtJ.~ 'P -- GU-S ~ v C oS e e 1vAe> -- -6IvIg (..M~e ~Ao 'PS;", e <..<)5 e DuAo = Tv Ao . Pc.o.s1.<9 'Psi"lec:...>s$J ~..L ~.s e Pc) =-fS'vt'° Gu.s (;) ':::' TC1V1 ::: e 1:'J '::' ~ (J~.S}(1.Sc» f:...o$7. ~7. 1;,'" 1.3 ':' = 1:5 ,:' c),v). 0 (~) fjo.,t--27.50 .. 7. '14 k:ps ES. ': f'v cry P = 7.;2.4 3.31 ~